keplers law of periods... AGAIN !!! (1 Viewer)

[asl2]

sorry guyz... it didn't work out... so heres the exat question and the exat data just like the books says:

"Use keplers law of periods to determine the time required for Mercury to complete an orbit around the Sun. Note that the radios of the Earth's orbit is 150 * 10^6 Km. RAdius of Mercury = 58.5 * 10 ^6 KM"

It does not mention anything about the sun...

the answer should be in earth years... so divide your answer by 60 which will give u minuites, then agian by 60, which will give hours, then by 24, which will gvie u days, then 365 which will give years...

the answer is 0.2**** something like this..

thx

 

[marsesbars]

1. If you have the value of 'M', the mass of the sun, then just plug it and 'r' into Kepler's Law to get 'T'. Simple.

2. If not, you can get 'M' by plugging into Kepler's Law the radius and period of earth's orbit around the sun. You know 'T' = 24 hours for earth (i hope ) and 'r' is given. Rearrange, get M. Then go to step 1.

 

[marsesbars]

A better method!

Just remembered a better method. You know the GM/(4pi^2) on the right hand side of the equation... Well, M is the mass of the thing being orbitted - which is the sun, for both Mercury and Earth - and everything else is constant, so the value of GM/(4pi^2) is the same for both planets

So.... you can equate the LHS's of the equations for Earth and Mercury to get

ratio (r cubed/ t squared) for mercury = ratio (r cubed/ t squared) for earth

Very quick

 

[kini mini]

Originally posted by marsesbars
1. If you have the value of 'M', the mass of the sun, then just plug it and 'r' into Kepler's Law to get 'T'. Simple.

2. If not, you can get 'M' by plugging into Kepler's Law the radius and period of earth's orbit around the sun. You know 'T' = 24 hours for earth (i hope ) and 'r' is given. Rearrange, get M. Then go to step 1.

T = 1 year for an orbit around the sun I think
I still think it is better to use the mass of the sun if you can, because there are less operations to do and less chance of making a mistake.

 

[OkDen]

bump
this physics sounds very interesting:d

necromancer